Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r - 3}{-r^2 + 2r + 63} \times \dfrac{r^2 - 3r - 54}{r + 6} $
Explanation: First factor out any common factors. $q = \dfrac{r - 3}{-(r^2 - 2r - 63)} \times \dfrac{r^2 - 3r - 54}{r + 6} $ Then factor the quadratic expressions. $q = \dfrac {r - 3} {-(r - 9)(r + 7)} \times \dfrac {(r - 9)(r + 6)} {r + 6} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {(r - 3) \times (r - 9)(r + 6) } { -(r - 9)(r + 7) \times (r + 6)} $ $q = \dfrac {(r - 9)(r + 6)(r - 3)} {-(r - 9)(r + 7)(r + 6)} $ Notice that $(r - 9)$ and $(r + 6)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {\cancel{(r - 9)}(r + 6)(r - 3)} {-\cancel{(r - 9)}(r + 7)(r + 6)} $ We are dividing by $r - 9$ , so $r - 9 \neq 0$ Therefore, $r \neq 9$ $q = \dfrac {\cancel{(r - 9)}\cancel{(r + 6)}(r - 3)} {-\cancel{(r - 9)}(r + 7)\cancel{(r + 6)}} $ We are dividing by $r + 6$ , so $r + 6 \neq 0$ Therefore, $r \neq -6$ $q = \dfrac {r - 3} {-(r + 7)} $ $ q = \dfrac{-(r - 3)}{r + 7}; r \neq 9; r \neq -6 $